What is the integral of (ln(x))/(x(3+2ln(x))) ?

The integral of (ln(x))/(x(3+2ln(x))) is 1/4 (3+2ln(x)-3ln|3+2ln(x)|)+C

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integral of (ln(x))/(x(3+2ln(x)))

\int \frac{ln ( x )}{x ( 3 + 2 ln ( x ) )} d x

\frac{1}{4} (3 + 2 ln (x) - 3 ln ∣ 3 + 2 ln (x) ∣) + C

Solution steps

\int \frac{ln ( x )}{x ( 3 + 2 ln ( x ) )} d x

Apply u-substitution

: \int \frac{u}{3 + 2 u} d u

= \int \frac{u}{3 + 2 u} d u

Apply u-substitution

: \int \frac{v - 3}{4 v} d v

= \int \frac{v - 3}{4 v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{v - 3}{v} d v

Expand

\frac{v - 3}{v} : 1 - \frac{3}{v}

= \frac{1}{4} \cdot \int 1 - \frac{3}{v} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{4} (\int 1 d v - \int \frac{3}{v} d v)

\int 1 d v = v

\int \frac{3}{v} d v = 3 ln ∣ v ∣

= \frac{1}{4} (v - 3 ln ∣ v ∣)

Substitute back

= \frac{1}{4} (3 + 2 ln (x) - 3 ln ∣ 3 + 2 ln (x) ∣)

Add a constant to the solution

= \frac{1}{4} (3 + 2 ln (x) - 3 ln ∣ 3 + 2 ln (x) ∣) + C

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What is the integral of (ln(x))/(x(3+2ln(x))) ?
The integral of (ln(x))/(x(3+2ln(x))) is 1/4 (3+2ln(x)-3ln|3+2ln(x)|)+C